The Second Derivative Test for Relative Maximum and Minimum - Problem 3
It is also possible to find relative maxima and minima by using the second derivative of a function. First, calculate the first derivative of the function and find its critical points. Then, calculate the second derivative of the function. Plug in the critical points found into the second derivative. If the result is positive, the graph is concave up at that point, so it is a relative minimum. If the result is negative, the graph is concave down at that point, so it is a relative maximum. If the second derivative is 0 at a critical point, then the second derivative test has failed, and you must use the first derivative test to determine if that point is a maximum or minimum.
It is also possible to find relative maxima and minima by using the second derivative of a function. First, calculate the first derivative of the function and find its critical points. Then, calculate the second derivative of the function. Plug in the critical points found into the second derivative. If the result is positive, the graph is concave up at that point, so it is a relative minimum. If the result is negative, the graph is concave down at that point, so it is a relative maximum.
If the second derivative is 0 at a critical point, then the second derivative test has failed, and you must use the first derivative test to determine if that point is a maximum or minimum.
We're using the second derivative test to find the relative maxima and minima of a function. Let's take a look at this function. It's a fourth degree function -½x to the fourth plus x³ plus 18x² minus 54x plus 108.
First we have to find the first and second derivatives. So let's find f'. and it's pretty straight forward. We're going to get 4 times -½, -2x³ plus 3x² plus 36x minus 54. The derivative of 108 is 0. That's the first derivative. Then the second derivative is going to be -6x² plus 6x plus 36. The derivative of -54 is 0. So here are my two derivatives, and I'm going to factor the second one. It looks a little easier to factor than the first one.
I'll pull a -6 out, and I have (x²-x-6). Then I see that this actually does factor. If I have what factors of -6 are going to give me a -1 in the middle. I think -3 and 2. (x-3)(x+2). This will give me -6. -3x plus 2x is -x. That works. So this is my second derivative all factored, and here is my first derivative.
The next thing I'm asked to do is find critical points. So it turns out that I need to factor that first derivative. Now that's not an easy thing to do. This is -2x³ plus 3x². Then we have the +36x minus 54. Now this to me looks like a candidate for factoring by grouping. The way that works is you take a look at the first two terms, and the last two. Factoring by grouping generally works on polynomials that have an even number of terms.
Here I can factor out of these two and x², and I'm left with -2x plus 3. Now if I can factor something out of this, and be left with a -2x plus 3 then this method works, and this will factor. Now I can factor out of this to get a -2x, I'd have to factor -18. Factoring a -18 out of 36 will leave a -2x. In fact I've got -18 out of 54 leaves a +3, that works.
So you see what I have here. I have an x² times -2x plus 3, and a -18 times -2x plus 3. So that's a common factor to both terms. I get x² minus 18 times -2x plus 3.
Now critical points are points where the derivative is 0, or undefined. This derivative is never going to be undefined. So we need only worry about where it's going to equal 0. It will equal 0 when this is 0, or when this is 0. Now this factor is 0 when x² equals 18. This one is 0 when -2x equals -3, so x equals 3/2. Here, x equals plus or minus root 18 which is the same as 3 root 2. This reduces to 3 root 2. So these are my three critical points; plus or -3 root 2, and 3/2.
So the last part is to compute the second derivative at each critical point. We want to see whether the second derivative is just positive or negative. We already know the first is 0 at each of these points.
So recall that the second derivative is -6(x-3)(x+2). So I have to test each of these critical points. Let's try the 3/2 first. I'll have -6 here. 3/2 minus 3. I don't even really need to know what the answer is, I just need to know whether it's positive or negative. So I'm going to skip the actual calculation and just write 3/2 minus 3, negative. Then I'm going to put 3/2 plus 2, positive. So it's negative times a negative times a positive. This is going to be greater than 0. That's all I need to know.
The fact that it's greater than 0 means that the curve is going to be concave up. It's got a horizontal tangent, and its concave up at 3/2. So we have a relative min at x equals 3/2
Now let's see what happens at 3 root 2. I need to do the same thing. I need to calculate f''(3 root 2). I get -6 times 3 root 2 minus 3 times 3 root 2 plus 2. Again I only need to know whether this is positive or negative. Keep in mind that 3 root 2 is bigger than the 3. This is 3 times 1.4. So this is positive, and this is positive. This is negative. So we're going to have a negative times a positive times a positive. This will all end up being negative.
Now before I conclude, let me do the same thing for the -3 root 2. It's going to be -6(-3 root 2 minus 3) (-3 root 2 plus 2). Here I have a negative. I have another negative, and this is going to be negative because -3 root 2 is definitely going to be smaller than 2, so we have another negative -3 which means I have a negative value here. Both of these second derivatives are negative. That means in both cases, the graph is going to be concave down with a horizontal tangent. So in both cases, I have a relative max. Relative max at x equals plus or minus 3 root 2.
That's how you use the second derivative test. Find critical points, and then test them in the second derivative. If you get a negative result, it means the graph's concave down. Then you're going to have a concave down right here. You're going to have a relative max. If the second derivative is positive, you're graphs concave up, and you have a relative min.